Question

Find the value of $A$ and $B$ if
$\cos (3A+B)=\frac{1}{2}$ and $\sin (3A-B)=\frac{1}{2}$

• A. $A={25}^{\circ }B={35}^{\circ }$
• B. $A={15}^{\circ }B={15}^{\circ }$
• C. $A=5^{\circ }B={18}^{\circ }$
• D. $A=5^{\circ }B={25}^{\circ }$

Answer 1

The correct option is B $A={15}^{\circ }B={15}^{\circ }$

Given that,

$\cos (3A+B)=\frac{1}{2}$ and $\sin (3A-B)=\frac{1}{2}$

To find out,

The values of $A$ and $B$.

We know that, $\cos {60}^o=\frac{1}{2}$

Hence, $\cos (3A+B)=\frac{1}{2}=\cos {60}^o$

$\Rightarrow 3A+B={60}^o....\left(1\right)$

We also know that, $\sin {30}^o=\frac{1}{2}$

Hence, $\sin (3A-B)=\frac{1}{2}=\sin {30}^o$

$\Rightarrow 3A-B={30}^o....\left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$, we get:

$3A+B+3A-B={90}^o$

$\Rightarrow 6A={90}^o$

$\Rightarrow A={15}^o$

Substituting the value of $A$ in $\left(2\right)$, we get:

$3\times {15}^o-B={30}^o$

$\Rightarrow B={45}^o-{30}^o$

$\Rightarrow B={15}^o$

Hence, $A={15}^o,B={15}^o$.